2788 + Interview Questions in GRAVITATION IN GENERAL AWARENESS Page 23 InterviewSolution

1101.	If the Earth losses its gravity, then for a bodyA. zero and sameB. same and zeroC. zero and zeroD. two times and zero
Answer» Correct Answer - B If the earth loses its gravity, then the mass of a body will remain the same but its weight (mg) will be zero. The answer is same and zero.

Discussion

1102.	If the Earth losses its gravity, then for a bodyA. Weight becomes zero, but not the massB. Mass becomes zero, but not the weightC. Both mass and weight become zeroD. Neither mass nor weight become zero
Answer» Correct Answer - A

Discussion

1103.	When escape velocity is given to a particle on surface of earth, its total energy isA. zeroB. greater than zeroC. less than zeroD. `-GMm//2R`
Answer» Correct Answer - A At `v lt v_(e)`, total energy is negative `v = v_(e)`, total energy is zero and `v gt v_(e)` total enegry is positive So, with the reference to above given cases, it can be said that when the escape velocity is given to a particle, its total energy is zero.

Discussion

1104.	If the Earth losses its gravity, then for a bodyA. weight becomes zero but not the massB. mass becomes zero but not weightC. neither mass nor weight is zeroD. both mass and weight are zero
Answer» Correct Answer - A As the earth loses its gravity then g=0 hence weight =mg =0 but the value of mass is unaffected or mass does not change

Discussion

1105.	Find the binding energy of a satellite of mass `m` in orbit of radius `r`, (R = radius of earth, g = acceleration due to gravity)A. `(mgR^(2))/(r )`B. `(mgR^(2))/(2r )`C. `-(mgR^(2))/(r )`D. `-(mgR^(2))/(2r )`
Answer» Correct Answer - B Binding energy is equal to negative value of total mechanical energy of a satellite the energy required to remove the satellite from its orbit around the earth to infinity is called binding energy of the satelite it is equal tonegative of total mechanical energy of satellite in its orbit thus binding energy `=-E=(GMm)/(2r)` but `g=(GM)/(R^(2))` `rarr GM=gR^(2)` or `BE=(gmR^(2))/(2r)`

Discussion

1106.	The binding energy of an object of mass `m` placed on the surface of the earth r is (R = radius of earth, g = acceleration due to gravity)A. `mg R//2`B. mg RC. `(mgR)/(4)`D. `(mgR)/(8)`
Answer» Correct Answer - B The energy required to remove an object from the surface of the earth to infinity is called binding energy of the object. It is equal to negative of gravitational potential energy of the object Thus, bindin energy `= -E=(GMm)/(R)` but, `g=(GM)/(R^(2))rArrGM=gR^(2)rArr BE=(gmR^(2))/(R)=gmR`.

Discussion

1107.	In a gravitational field, if a body is bound with earth, then total mechanical energy isA. positiveB. zeroC. negativeD. may be positive, negative or zero
Answer» Correct Answer - C Total mechanical energy of any closed system is always negative.

Discussion

1108.	Assertion : Gravitational potential of earth at every place body are equal to one. Reason: Everybody on earth is bound by the attraction of earth.A. If both the assertion and reason are true and reason is a true explantion of the assertion.B. If both the assertion and reason are true but the reason is not true the correct explantion of the assertion.C. If the assertion is true but reason falseD. If both the assertion and reason are false.
Answer» Correct Answer - A Becuase gravitational force is always attractive in nature and everybody is bonded by this gravitational force of attraction of earth.

Discussion

1109.	Out of the following interactions, weakest isA. gravitationalB. electromagneticC. nuclearD. electrostatic
Answer» Correct Answer - A

Discussion

1110.	If the earth satellite is put into an orbit at a height where resistance due to atmosphere, cannot be neglected, how will motion of satellite be affected?
Answer» Air resistance will reduce the orbital velocity of satellite. Hence, the satellite will ultimately fall back to the earth. Air resistance may also produce a lot of heat and the satellite may even burn.

Discussion

1111.	Which of the following interaction is the weakestA. GravitationalB. ElectrostaticC. NuclearD. Electromagnetic
Answer» Correct Answer - a

Discussion

1112.	An artificial satellite of the earth releases a packet. If air resistance is neglected, the point where the packet will hit, will beA. aheadB. exactely belowC. behindD. it will never reach the earth
Answer» Correct Answer - D

Discussion

1113.	If the earth suddenly loses its power of attraction, then a body on the surface of the earth will beA. reduced to its mass zeroB. reduced to its weight zeroC. both mass and weight reduced to zeroD. can not be predicted
Answer» Correct Answer - b

Discussion

1114.	The value of G depends uponA. the masses of bodiesB. the medium between the bodiesC. the temperature of bodiesD. system of units
Answer» Correct Answer - d

Discussion

1115.	The orbital velocity of a satellite at point `B` with radius `r_(B)` and `n`. The radius of a point `A` is `r_(A)`. If the orbit is increased in radial distance so that `r_(A)` becomes `.12r_(A)` find the orbital velocity at `(1.2 r_(A))`: A. `(vr_(B))/(r_(A)sqrt(1.2))`B. `(vr_(A))/(1.2r_(B))`C. `(vr_(B))/(1.2r_(A))`D. `(vr_(A))/(r_(B)sqrt(2))`
Answer» Correct Answer - A For two points on same orbit `L=mv_(A)r_(A)=mvr_(B)` `v_(A)=(vR_(B))/(r_(A))` For two points on different orbits. ` v=sqrt((GM)/r) (v_(0))/(v_(A))=((r_(A))/(1.2r_(A)))^(1//2)` `v_(0)=v_(A) ((r_(A))/(1.2r_(A)))^(1//2)=(vr_(B))/(r_(A))((r_(A))/(1.2r_(A)))^(1//2)=(vr_(B))/(r_(A) sqrt(1.2))`

Discussion

1116.	A satellite is moving in a circular orbit round the earth. If any other planet comes in between them, it willA. continue to move with the same speed along the same pathB. move with the same velocity tangential to original orbit.C. Fall down with increasing velocity.D. come to rest after moving certain distance along original path.
Answer» Correct Answer - B

Discussion

1117.	When a satellite is orbitting round a planet in a circular orbit, work done y the gravitationaol force acting on the satellite isA. zero on completing one revolution onlyB. zero alwaysC. infiniteD. negative
Answer» Correct Answer - b

Discussion

1118.	If radius of the earth contracts to half of its present value without change in its mass, what will be the new duration of the day?A. `12hr`B. `6hr`C. `42 mi n`D. none of these
Answer» Correct Answer - B `I=2/5MR^(2)` `L=Iomega`=constant `T=(2pi)/(omega)implies T prop1/(omega) prop I prop R^(2)` `T prop R^(2)` So if radius becomes half, `T` becomes `1/4` times i.e. `6` hrs.

Discussion

1119.	The gravitational potential energy of a two-particle system is derived in this chapter as U = -Gm₁m₂/r. Does it follow from the equation that the potential energy for r = ∞ must be zero? Can we choose the potential energy for r = ∞ to be 20 J and still use this formula? If no, what formula should be used to calculate the gravitational potential energy at separation r?
Answer» Yes. Since U is inversely proportional to r, at r = ∞ the gravitational potential energy must be zero. If we choose the potential energy at r = ∞ to be 20 J, we can not use this formula, because putting r = ∞, will not give the result 20 J. The required formula will be, U = -Gm_₁m_₂/r + 20.

1119.

The gravitational potential energy of a two-particle system is derived in this chapter as U = -Gm₁m₂/r. Does it follow from the equation that the potential energy for r = ∞ must be zero? Can we choose the potential energy for r = ∞ to be 20 J and still use this formula? If no, what formula should be used to calculate the gravitational potential energy at separation r?

Answer»

Yes. Since U is inversely proportional to r, at r = ∞ the gravitational potential energy must be zero.

If we choose the potential energy at r = ∞ to be 20 J, we can not use this formula, because putting r = ∞, will not give the result 20 J.

The required formula will be, U = -Gm_₁m_₂/r + 20.

Discussion

1120.	Suppose the gravitational potential due to a small system is k/r² at a distance r from it. What will be the gravitational field? Can you think of any such system? What happens if there were negative masses?
Answer» The gravitational potential V = k/r². Hence gravitational field E = - dV/dr = -(-2k/r³) = 2k/r³ A hypothetical gravitational dipole system can be assumed like this, as in the case of an electric dipole in which the electric field is inversely proportional to the cube of the distance. It is hypothetical because in the case of gravitation negative masses do not occur. If there were negative masses the direction of the field would be reversed.

1120.

Suppose the gravitational potential due to a small system is k/r² at a distance r from it. What will be the gravitational field? Can you think of any such system? What happens if there were negative masses?

Answer»

The gravitational potential V = k/r².

Hence gravitational field E = - dV/dr = -(-2k/r³) = 2k/r³

A hypothetical gravitational dipole system can be assumed like this, as in the case of an electric dipole in which the electric field is inversely proportional to the cube of the distance. It is hypothetical because in the case of gravitation negative masses do not occur.

If there were negative masses the direction of the field would be reversed.

Discussion

1121.	A stone is dropped from a height of 20m. a) How long will it take to reach the ground? b) What will be its speed when it hits the ground?
Answer» a) Using the relation: s = ut + ½ gt² t² = 20/5 = 4 t = 2s b) For a freely falling body: v² = u² + 2gh v² = 400 v = 20 m/s

1121.

A stone is dropped from a height of 20m. a) How long will it take to reach the ground? b) What will be its speed when it hits the ground?

Answer»

a) Using the relation:

s = ut + ½ gt²

t² = 20/5 = 4

t = 2s

b) For a freely falling body:

v² = u² + 2gh

v² = 400

v = 20 m/s

Discussion

1122.	If the value of `g` at the surface of the earth is `9.8 m//sec^(2)`, then the value of `g` at a place `480` km above the surface of the earth will be (Radius of the earth is `6400` km)A. `8.4 m//sec^(2)`B. `9.8 m//sec^(2)`C. `7.2 m//sec^(2)`D. `4.2 m//sec^(2)`
Answer» Correct Answer - A

Discussion

1123.	Which of the following graph depicts relation between time period (T) and radius of orbit (r) of a planet ?A. B. C. D.
Answer» Correct Answer - a `T^(2)=((4pi)/(GM))^(4)r^(3)` `y="m x"` Graph between `T^(2)` and `r^(3)` is a straight line passing through origin.

Discussion

1124.	As it falls, the acceleration of a body dropped from the height equal to that of radius of earth,A. remains the sameB. decreasesC. increasesD. initially increases then decreases
Answer» Correct Answer - C As the height above the surface of earth decreases, the acceleration due to gravity increases.

Discussion

1125.	In the relation `F=G M m//d^(2)`, the quantity GA. depends on the value of g at the place of observationB. is used only when the earth is one of the two massesC. is greatest at the surface of the earthD. is universal constant of nature
Answer» Correct Answer - D In the relation `F=(GMm)/d^(2)`, the quality G is universal gravitational constant of nature.

Discussion

1126.	The value of acceleration due to gravity on the Earth at a distance of 29,000 km from the surgace is `0.3 ms^(-2)`. The value of acceleration due to gravity at the same height on a planet whose mass is `66.70 xx 10^(22)` kg and diameter is 8700 km is _______ `ms_(-2)`. `("Take" G = 6.67 xx 10^(-11) N m^(2) kg^(-2))`A. 0.05B. 0.04C. 0.06D. 0.09
Answer» Correct Answer - B `g = (GM)/(R + h)^(2) = (6.67 xx 10^(-11) xx 66.7 xx 10^(22))/([(4350 + 29000) xx 10^(3)]^(2))` `= (444.889 xx 10^(11))/((33350)^(2)xx10^(6))=0.04 m s^(-2)`

Discussion

1127.	The value of acceleration due to gravityA. is same on equator and polesB. is least on polesC. is least on equatorD. increase from pole to equator
Answer» Correct Answer - C The value of acceleration due to gravity is least on equater, as equatorial radius is maximum.

Discussion

1128.	A boy is whirling a stone tied with a string in a horizontal circular path. When the string breacks, the stoneA. Will continue to move in the circular pathB. will move alonge a straight line towards the circular pathC. will move along a straight line tangential to the circular pathD. will move along a straight line perpendicular to the circular path away from the boy
Answer» Correct Answer - C When the string breaks,the stone will move along a st. line tangential to the circular path. This is because centripetal force required for moving in circular path is no longer provide.

Discussion

1129.	The ratio of the masses of two planets is 1 : 10 and the ratio of their diameters is 1 : 2. If the length of a seconds pendulum on the first planet is 0.4 m, then the length of the seconds pendulum on the second planet is _______.A. 10 cmB. 0.5 mC. 10 mD. 1.0 m
Answer» Correct Answer - D `m_(1) : m_(2) = 1 : 10` `d_(1) : d_(2) = 1 : 2 rArr r_(1) : r_(2) = 1 : 2` `l_(1) = 0.4 m` `l_(2) = ?` `g_(1) = (GM_(1))/(R_(1)^(2)), g_(2) = (GM_(2))/(R_(2)^(2)` `(g_(1))/(g_(2))=(M_(1))/(R_(1)^(2)) xx (R_(2)^(2))/(M_(2)) = (1)/(10)xx(2^(2))/(1^(2))=(4)/(10)` `(T_(1))/(T_(2))=sqrt(l_(1)/(g_(1))xx(g_(2))/(l_(2)))=sqrt((0.4)/(l_(2))xx(10)/(4))` `(2)/(2) = sqrt((1)/(l_(2))),(1)/(l_(2)) = (1)/(1) rArr l_(2) = 1m`

Discussion

1130.	Find the value of acceleration due to gravity at the surface of moon whose mass is `7.4xx10^(22)` kg and its radius is `1.74xx10^(22)` kg
Answer» Correct Answer - `1.6 m//s^(2)` Hint : Use `g=(GM)/(R^(2))`

Discussion

1131.	If the ratio of the masses of two planets is 8 : 3 and the ratio of their diameters is 2 : 3, then what will be the ratio of their acceleration due to gravity ?
Answer» Correct Answer - `6 : 1` Hint : By `g=(GM)/(R^(2))`

Discussion

1132.	Whathat will be the acceleration due to gravity on a planet whose mass is 4 times that of earth and identical in size ?
Answer» Correct Answer - 4 times that of earth Hint : Use `g=(GM)/(R^(2))`

Discussion

1133.	The graph that represents the relation between orbital velocity `(v_(0))` of a satellite and radius (r) of the orbit around earth isA. B. C. D.
Answer» Correct Answer - a `v_(0)=sqrt((GM)/(r)),v_(0)^(2) prop (1)/(r)` `Y^(2)x` = constant. It is a curve.

Discussion

1134.	The acceleration due to gravity g on the earth is `9.8m//s^(2)`. What would be the value of g for a planet whose size is the same as that of earth and the density is twice that of earth?A. `19.6m//s^(2)`B. `9.8m//s^(2)`C. `4.9m//s^(2)`D. `2.45m//s^(2)`
Answer» Correct Answer - a `g_(1) prop rho_(1)" and "g_(2) prop rho_(2)`

Discussion

1135.	What are the values of the escape velocities for the moon and the sun respectively?
Answer» They are, V_moon = 2.4 km s^-1 and V_sun = 620 km s^-1.

Discussion

1136.	From where does a satellite revolving around a planet get the required centripetal force?
Answer» From the gravitational attraction of planet exerted on satellite.

Discussion

1137.	State the relationship between 'g' and 'G'?
Answer» The relationship between 'g' and 'G' is: g = GM_e/R_e²

1137.

State the relationship between 'g' and 'G'?

Answer»

The relationship between 'g' and 'G' is:

g = GM_e/R_e²

Discussion

1138.	What is the value of the escape velocity on the surface of the earth?
Answer» It is 11.2 km s^-1

Discussion

1139.	What is meant by orbital velocity?
Answer» The velocity that must be imparted to a body to keep it in an orbit round the earth is known as its orbital velocity.

Discussion

1140.	What is meant by escape velocity?
Answer» The velocity with which a body must be projected in order that it may escape from the earth's atmosphere (gravitation), is called escape velocity.

Discussion

1141.	The value of the gravitational acceleration at the height h to be `1%` of its value at the surface of earth, then h is equal toA. 6400 kmB. 57,600 kmC. 2560 kmD. 64,000 km
Answer» Correct Answer - b `g_(h)=g((R)/(R+h))^(2)`

Discussion

1142.	The centripetal acceleration of a satellite that circles the earth at an altitude 400 km above see level is (g on the surface of earth is `10m//s^(2)`, Radius of the earth is `6.4xx10^(6)m`)A. `8.75m//s^(2)`B. `9.2m//s^(2)`C. `10m//s^(2)`D. `7.5m//s^(2)`
Answer» Correct Answer - a `g_(2)=g[1-(2h)/(R)]=10[1-(2xx400)/(6400)]` `=10[1-(1)/(8)]` `=10[(8-1)/(8)]=(70)/(8)=8.75m//s^(2).`

Discussion

1143.	If g is acceleration due to gravity at the surface of the earth, then its value at a depth of 1/4 of the radius of the earth is
Answer» Correct Answer - d `g_(d)=g(1-(d)/(R))=g(1-(R)/(4R))=(3)/(4)g.`

Discussion

1144.	The acceleration due to gravity decreases by `Deltag_(1)` when a body is taken to a small height `h lt lt R.` The acceleration due to gravity decreases by `Deltag_(2)` when the body is taken to a depth h from the surface off the earth, then (R= Radius of the earth)A. `Deltag_(1)=Deltag_(2)`B. `Deltag_(1)=2Deltag_(2)`C. `Deltag_(2)=2Deltag_(1)`D. `Deltag_(1)=4Deltag_(2)`
Answer» Correct Answer - b `g_(h)=g(1-(2h)/(R))` `Deltag_(1)=g-g_(h)=(2hg)/(R)` `g_(d)=g(1-(h)/(R))` `Deltag_(2)=g-d_(d)=(gh)/(R)` `therefore" "Deltag_(2)=2Deltag_(2).`

Discussion

1145.	In the above quation, velocity with which the sone hits the ground isA. `39.2m//s`B. `19.6m//s`C. zeroD. `78.4m//s`
Answer» Correct Answer - A `v=u+at=0+9.8xx4=39.2m//s`

Discussion

1146.	What is the direction of ‘g’? Throw a stone vertically up. Measure the time required for it to come back to earth’s surface with stop clock.a) What happens to its speed when it moves up and down?b) What is the direction of acceleration?
Answer» a) When the stone moves up, its velocity decreases gradually and finally it becomes zero, because it is moving against the direction of gravitation. When its velocity becomes zero, it starts to fall down due to gravitational force of earth, and its velocity increases. b) While the stone is moving up, the direction of acceleration is in upward direction, which is against to the direction of gravitational force. While the stone is falling down, the direction of acceleration is in downward direction, which is in the direction of gravitational force.

1146.

What is the direction of ‘g’? Throw a stone vertically up. Measure the time required for it to come back to earth’s surface with stop clock.a) What happens to its speed when it moves up and down?b) What is the direction of acceleration?

Answer»

a)

When the stone moves up, its velocity decreases gradually and finally it becomes zero, because it is moving against the direction of gravitation.
When its velocity becomes zero, it starts to fall down due to gravitational force of earth, and its velocity increases.

b)

While the stone is moving up, the direction of acceleration is in upward direction, which is against to the direction of gravitational force.
While the stone is falling down, the direction of acceleration is in downward direction, which is in the direction of gravitational force.

Discussion

1147.	The graph that represents variation of g with height (h) from the surface of the earth isA. B. C. D.
Answer» Correct Answer - a `g_(h)=g((R)/(R+h))^(2)`. It is a curve.

Discussion

1148.	Find the gravitational force between two masses 15 kg each separated by a distance of 2m. (G = 6.67 × 10-11Nm2kg-2 )
Answer» Given, m₁ = m₂ = 15 kg d = 2 m G = 6.67 × 10^-11Nm²kg^-2 ∴ F = \(\frac {m_1m_2}{d^2}\) \(= \frac {6.67 \times 10^{-11} \times 15 \times 15}{2^2}\) = 3.75 × 10^-9N.

1148.

Find the gravitational force between two masses 15 kg each separated by a distance of 2m. (G = 6.67 × 10-11Nm2kg-2 )

Answer»

Given, m₁ = m₂ = 15 kg

d = 2 m

G = 6.67 × 10^-11Nm²kg^-2

∴ F = \(\frac {m_1m_2}{d^2}\)

\(= \frac {6.67 \times 10^{-11} \times 15 \times 15}{2^2}\)

= 3.75 × 10^-9N.

Discussion

1149.	The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth isA. 0.15B. 0.04C. 1D. 6
Answer» Correct Answer - A

Discussion

1150.	The mass and weight of an object on Earth is 5kg and 49 N respectively. What will be their values on the Moon? Assume that the acceleration due to gravity on the Moon is `1/6` th of that on the Earth.
Answer» Given: `m_(e)=5kg` `W_(e)=49N` `g_(m)=1/6 g_(e)=1/6xx9.8=1.633m//s^(2)` To find: Mass of Moon `m_(m)=?` Weight on Moon `W_(m)=?` formula `W=F=mg` Solution: Mass remais same: `m_(m)=5kg` `W_(m)=m_(m)xxg_(m)` `=5xx1.633` `W_(m)=8.17N` Mass on Moon is 5 kg and weight is `8.17N`

Discussion

Explore topic-wise InterviewSolutions in .

If the Earth losses its gravity, then for a bodyA. zero and sameB. same and zeroC. zero and zeroD. two times and zero

If the Earth losses its gravity, then for a bodyA. Weight becomes zero, but not the massB. Mass becomes zero, but not the weightC. Both mass and weight become zeroD. Neither mass nor weight become zero

When escape velocity is given to a particle on surface of earth, its total energy isA. zeroB. greater than zeroC. less than zeroD. `-GMm//2R`

If the Earth losses its gravity, then for a bodyA. weight becomes zero but not the massB. mass becomes zero but not weightC. neither mass nor weight is zeroD. both mass and weight are zero

Find the binding energy of a satellite of mass `m` in orbit of radius `r`, (R = radius of earth, g = acceleration due to gravity)A. `(mgR^(2))/(r )`B. `(mgR^(2))/(2r )`C. `-(mgR^(2))/(r )`D. `-(mgR^(2))/(2r )`

The binding energy of an object of mass `m` placed on the surface of the earth r is (R = radius of earth, g = acceleration due to gravity)A. `mg R//2`B. mg RC. `(mgR)/(4)`D. `(mgR)/(8)`

In a gravitational field, if a body is bound with earth, then total mechanical energy isA. positiveB. zeroC. negativeD. may be positive, negative or zero

Out of the following interactions, weakest isA. gravitationalB. electromagneticC. nuclearD. electrostatic

If the earth satellite is put into an orbit at a height where resistance due to atmosphere, cannot be neglected, how will motion of satellite be affected?

Which of the following interaction is the weakestA. GravitationalB. ElectrostaticC. NuclearD. Electromagnetic

An artificial satellite of the earth releases a packet. If air resistance is neglected, the point where the packet will hit, will beA. aheadB. exactely belowC. behindD. it will never reach the earth

If the earth suddenly loses its power of attraction, then a body on the surface of the earth will beA. reduced to its mass zeroB. reduced to its weight zeroC. both mass and weight reduced to zeroD. can not be predicted

The value of G depends uponA. the masses of bodiesB. the medium between the bodiesC. the temperature of bodiesD. system of units

When a satellite is orbitting round a planet in a circular orbit, work done y the gravitationaol force acting on the satellite isA. zero on completing one revolution onlyB. zero alwaysC. infiniteD. negative

If radius of the earth contracts to half of its present value without change in its mass, what will be the new duration of the day?A. `12hr`B. `6hr`C. `42 mi n`D. none of these

Suppose the gravitational potential due to a small system is k/r² at a distance r from it. What will be the gravitational field? Can you think of any such system? What happens if there were negative masses?

A stone is dropped from a height of 20m. a) How long will it take to reach the ground? b) What will be its speed when it hits the ground?

If the value of `g` at the surface of the earth is `9.8 m//sec^(2)`, then the value of `g` at a place `480` km above the surface of the earth will be (Radius of the earth is `6400` km)A. `8.4 m//sec^(2)`B. `9.8 m//sec^(2)`C. `7.2 m//sec^(2)`D. `4.2 m//sec^(2)`

Which of the following graph depicts relation between time period (T) and radius of orbit (r) of a planet ?A. B. C. D.

As it falls, the acceleration of a body dropped from the height equal to that of radius of earth,A. remains the sameB. decreasesC. increasesD. initially increases then decreases

In the relation `F=G M m//d^(2)`, the quantity GA. depends on the value of g at the place of observationB. is used only when the earth is one of the two massesC. is greatest at the surface of the earthD. is universal constant of nature

The value of acceleration due to gravityA. is same on equator and polesB. is least on polesC. is least on equatorD. increase from pole to equator

The ratio of the masses of two planets is 1 : 10 and the ratio of their diameters is 1 : 2. If the length of a seconds pendulum on the first planet is 0.4 m, then the length of the seconds pendulum on the second planet is _______.A. 10 cmB. 0.5 mC. 10 mD. 1.0 m

Find the value of acceleration due to gravity at the surface of moon whose mass is `7.4xx10^(22)` kg and its radius is `1.74xx10^(22)` kg

If the ratio of the masses of two planets is 8 : 3 and the ratio of their diameters is 2 : 3, then what will be the ratio of their acceleration due to gravity ?

Whathat will be the acceleration due to gravity on a planet whose mass is 4 times that of earth and identical in size ?

The graph that represents the relation between orbital velocity `(v_(0))` of a satellite and radius (r) of the orbit around earth isA. B. C. D.

The acceleration due to gravity g on the earth is `9.8m//s^(2)`. What would be the value of g for a planet whose size is the same as that of earth and the density is twice that of earth?A. `19.6m//s^(2)`B. `9.8m//s^(2)`C. `4.9m//s^(2)`D. `2.45m//s^(2)`

What are the values of the escape velocities for the moon and the sun respectively?

From where does a satellite revolving around a planet get the required centripetal force?

State the relationship between 'g' and 'G'?

What is the value of the escape velocity on the surface of the earth?

What is meant by orbital velocity?

What is meant by escape velocity?

The value of the gravitational acceleration at the height h to be `1%` of its value at the surface of earth, then h is equal toA. 6400 kmB. 57,600 kmC. 2560 kmD. 64,000 km

The centripetal acceleration of a satellite that circles the earth at an altitude 400 km above see level is (g on the surface of earth is `10m//s^(2)`, Radius of the earth is `6.4xx10^(6)m`)A. `8.75m//s^(2)`B. `9.2m//s^(2)`C. `10m//s^(2)`D. `7.5m//s^(2)`

If g is acceleration due to gravity at the surface of the earth, then its value at a depth of 1/4 of the radius of the earth is

In the above quation, velocity with which the sone hits the ground isA. `39.2m//s`B. `19.6m//s`C. zeroD. `78.4m//s`

What is the direction of ‘g’? Throw a stone vertically up. Measure the time required for it to come back to earth’s surface with stop clock.a) What happens to its speed when it moves up and down?b) What is the direction of acceleration?

The graph that represents variation of g with height (h) from the surface of the earth isA. B. C. D.

Find the gravitational force between two masses 15 kg each separated by a distance of 2m. (G = 6.67 × 10-11Nm2kg-2 )

The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the acceleration due to gravity at the surface of the moon to that at the surface of the earth isA. 0.15B. 0.04C. 1D. 6

The mass and weight of an object on Earth is 5kg and 49 N respectively. What will be their values on the Moon? Assume that the acceleration due to gravity on the Moon is `1/6` th of that on the Earth.